Calculate de Broglie wavelength associated with an electron, accelerated through a potential difference of `400 V`.

Options

A. (a)0.03 nm
B. (b)0.04 nm
C. (c)0.12 nm
D. (d)0.06 nm

Answer

To calculate the de Broglie wavelength associated with an electron that has been accelerated through a potential difference of 400 V, we can follow these steps:

### Step 1: Understand the relationship between potential difference and kinetic energy
When an electron is accelerated through a potential difference (V), it gains kinetic energy (K.E.) equal to the charge of the electron multiplied by the potential difference. The formula for kinetic energy in electron volts (eV) is:
\[ K.E. = e \cdot V \]
Where:
- \( e \) (the charge of the electron) is approximately \( 1.6 \times 10^{-19} \) coulombs.
- \( V \) is the potential difference in volts.

For \( V = 400 \, \text{V} \):
\[ K.E. = 1.6 \times 10^{-19} \, \text{C} \times 400 \, \text{V} = 6.4 \times 10^{-17} \, \text{J} \]

### Step 2: Use the de Broglie wavelength formula
The de Broglie wavelength (\( \lambda \)) is given by the formula:
\[ \lambda = \frac{h}{p} \]
Where:
- \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \, \text{Js} \)).
- \( p \) is the momentum of the electron.

### Step 3: Relate momentum to kinetic energy
The momentum (\( p \)) of the electron can be expressed in terms of its kinetic energy:
\[ p = \sqrt{2mK.E.} \]
Where:
- \( m \) is the mass of the electron (\( 9.1 \times 10^{-31} \, \text{kg} \)).

### Step 4: Substitute kinetic energy into the momentum formula
Substituting \( K.E. \) into the momentum formula:
\[ p = \sqrt{2 \cdot (9.1 \times 10^{-31} \, \text{kg}) \cdot (6.4 \times 10^{-17} \, \text{J})} \]

### Step 5: Calculate the momentum
Calculating the value:
\[ p = \sqrt{2 \cdot 9.1 \times 10^{-31} \cdot 6.4 \times 10^{-17}} \approx \sqrt{1.16288 \times 10^{-46}} \approx 1.08 \times 10^{-23} \, \text{kg m/s} \]

### Step 6: Calculate the de Broglie wavelength
Now, substituting the value of momentum back into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34} \, \text{Js}}{1.08 \times 10^{-23} \, \text{kg m/s}} \approx 6.14 \times 10^{-11} \, \text{m} \]

### Step 7: Convert to nanometers
To convert meters to nanometers:
\[ \lambda \approx 0.0614 \, \text{nm} \text{ (or } 61.4 \, \text{pm)} \]

### Final Answer
The de Broglie wavelength associated with the electron is approximately \( 0.0614 \, \text{nm} \).

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